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Old 2019-05-05, 21:14   #1
hansl
 
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Default Observations of Wieferich primes and Wieferich-1 friendly club

This is a discussion on the forms of numbers found here: http://oeis.org/A239875
Quote:
Friendly club of (first Wieferich prime)-1 and (second Wieferich prime)-1: numbers n such that sigma(n)/n = 112/39.
It seems remarkable that the only two known Wieferich primes are friendly in this way.

It is noted on that page that numbers of the form (39 * perfect numbers) also follow this abundancy index.

One thing to test is the primality of these (39 * perfect numbers)+1:
(39*(2^(3-1))*((2^3)-1))+1 = 1093 (one of the known Wieferich primes)
(39*(2^(31-1))*((2^31)-1))+1 = 89927877317458132993 (prime but not Wieferich prime)

I am attempting to test the remaining perfect numbers based on currently known mersenne primes, using PFGW:
Code:
ABC (39*(2^($a-1))*((2^$a)-1))+1
2
3
5
7
13
17
19
31
61
89
107
127
521
607
1279
2203
2281
3217
4253
4423
9689
9941
11213
19937
21701
23209
44497
86243
110503
132049
216091
756839
859433
1257787
1398269
2976221
3021377
6972593
13466917
20996011
24036583
25964951
30402457
32582657
37156667
42643801
43112609
57885161
74207281
77232917
82589933
So far I have checked all up to 2976221, which is currently being crunched on. No primes besides the two above have been found.
Has this been checked before?

One other observation/conjecture I have made, is regarding the prime factors of (Wieferich primes-1) and friends. It seems that they are composed entirely of mersenne prime exponents, or mersenne primes themselves. This is trivially true for 39*perfect number form, but also holds for the two known numbers in the sequence which do not follow that form:
a(2)=3510 = 2^1 * 3^3 * 5^1 * 13^1 (2,3,5,13 are Mp exponents)
a(5)=11504844 = 2^2 * 3^2 * 13^2 * 31^1 * 61^1
In this case the factor 31 could be viewed as either the exponent of M31, or M5 itself.
Similarly the factor 3 could be considered an exponent of M3, or M2 itself.

So with this taken into account, it may be that a mersenne prime factor always occurs exactly once for these numbers, with the remaining factors being entirely mersenne prime exponents.

Any ideas if some form of this conjecture could be proven? I haven't seen any mention of such an idea before, but maybe its been thought of already, or that it follows somewhat easily from some related theorems?

Another note from the oeis page on Wieferich primes themselves: http://oeis.org/A001220
Quote:
In a 1977 paper, Wells Johnson, citing a suggestion from Lawrence Washington, pointed out the repetitions in the binary representations of the numbers which are one less than the two known Wieferich primes; i.e., 1092 = 10001000100 (base 2); 3510 = 110110110110 (base 2). It is perhaps worth remarking that 1092 = 444 (base 16) and 3510 = 6666 (base 8), so that these numbers are small multiples of repunits in the respective bases. Whether this is mathematically significant does not appear to be known. - John Blythe Dobson, Sep 29 2007
I have a more general(and possibly even more dubious) observation which is: they can also be represented as what I call pseudo-palindromes in base 2 (palindrome with trailing zeroes, or palindrome times a power of two). Many, but not all this friendly club also have this trait:
Code:
Base 2 conversions:
    1092 = 10001000100 (palindrome * 2^2)
    3510 = 110110110110 (palindrome * 2^1)
    19344 = 100101110010000 (not pseudo-palindrome)
    316992 = 1001101011001000000 (palindrome * 2^6)
    11504844 = 101011111000110011001100 (not pseudo-palindrome)
    1308463104 = 1001101111111011001000000000000 (palindrome * 2^12)
    335004893184 = 100110111111111110110010000000000000000 = (palindrome * 2^16)
    5360108961792 = 1001101111111111111011001000000000000000000 = (palindrome * 2^18)
    89927877317458132992 = 1001101111111111111111111111111011001000000000000000000000000000000 = (palindrome * 2^30)
    103679783671223438041533012022199844864 = 1001101111111111111111111111111111111111111111111111111111111011001000000000000000000000000000000000000000000000000000000000000 (palindrome * 2^60)
Does anyone think this could have any significance?
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Old 2019-05-05, 21:45   #2
GP2
 
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I think maybe the Wieferich@Home project tested repeating bit patterns to try to find a new Wieferich prime. Or maybe they just exhaustively tested all values, I don't know.

They didn't find one, and now that project is defunct and the web domain has been taken over by someone else for a different purpose. Here is an archived stats page from 2016 from the Internet archive.

Have you checked if the odd coincidences (bit patterns and friendly patterns) of the two Wieferich primes are also applicable in some way to the Base-a Wieferich primes for bases other than 2?



PS, needless to say, if any factor of a Mersenne number with prime exponent turns out to be non-squarefree, it would automatically become a new Wieferich prime. I regularly check newly discovered factors, but obviously the odds are astronomical.

The non-base-2 Wieferich primes do have some cases that correspond to non-squarefree factors. For example:

Code:
3^5−1 = 11^2 * ...
7^2+1 = 5^2 * ...
13^431+1 = 863^2 * ...
17^24473−1 = 48947^2 * ...
18^2+1 = 5^2 * ...
18^3−1 = 7^3 * ...
18^214007+1 = 1284043^2 * ...
19^3+1 = 7^3 * ...
22^3−1 = 13^2 * ...
22^246183293−1 = 492366587^2 * ...
22^44999368331−1 = 9809862296159^2 * ...
24^2−1 = 5^2 * ...
26^2−1 = 5^2 * ...
26^7+1 = 71^2 * ...
26^3347628353+1 = 6695256707^2 * ...
28^11+1 = 23^2 * ...
30^3−1 = 7^2 * ...
Even the two Wieferich primes 1093 and 3511 show up as non-squarefree factors for bases b that are powers of two. For example:

(2^14)^13 + 1 is divided by 1093^2
(2^26)^7 + 1 is divided by 1093^2
(2^28)^13 − 1 is divided by 1093^2

(2^135)^13 − 1 is divided by 3511^2
(2^270)^13 − 1 is divided by 3511^2
(2^351)^5 − 1 is divided by 3511^2

But of course, it's no big deal if a Mersenne number with composite exponent has a non-squarefree factor.
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Old 2019-05-11, 18:26   #3
hansl
 
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Quote:
Originally Posted by GP2 View Post
Have you checked if the odd coincidences (bit patterns and friendly patterns) of the two Wieferich primes are also applicable in some way to the Base-a Wieferich primes for bases other than 2?
I started looking into this, but I have difficulty in defining something analogous to mersenne numbers, perfect numbers, and abundancy index for any general base other than 2.

I did stumble upon the definition of k-hyperperfect numbers, which seems to be a reasonable generalization of such a thing (where k=base-1).
For base 3 as an example, there are 2-hyperperfect numbers: https://oeis.org/A007593

Quote:
Originally Posted by A007593
The known examples are all of the form 3^(k-1)*(3^k-2), where 3^k-2 is prime (cf. A014224). Conversely, from sigma(3^(k-1)*p)=(3^k-1)/2*(p+1) it is immediate that 2*sigma(n)=3n+1 for such numbers, i.e., they are 2-hyperperfect.
So based on that, it seems that a reasonable mersenne number definition for general bases would be: b^n-(b-1)

However this lacks the property that all primes of this form have a prime exponent.

Another generalization of mersenne numbers(primes) could be:
(b^n-1)/(b-1)
(AKA repunits in base b)
This actually does seem to have the property where only prime exponents produce primes.

So mersenne numbers having base 2 are unique in this case because its the only base where these two generalized forms are actually equivalent: 2^n-(2-1) == (2^n-1)/(2-1)

It may be that (if there is any relevance to the patterns in my OP at all), that they only apply to base 2 wieferich's.

One last thing which I tried to understand, but got a bit stuck, is if there is any way to define a "k-hyperabundancy index", or "k-hyperfriendlyness"
Such that all k-hyperperfect numbers are k-hyperfriendly for a given k, and also where this generalized formula is equivalent to the regular abundancy index given k=1.

Edit: I should also note, that for the case of "pseudopalindromes", I did not see any such patterns for wieferich primes of other prime bases, (base 3, or base 5 for example).

Last fiddled with by hansl on 2019-05-11 at 18:40 Reason: palindrome note
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Old 2020-09-02, 10:40   #4
JeppeSN
 
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I wrote that OEIS entry. Did you ever complete that ABC file to 2976221 or beyond? /JeppeSN
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